(c)

We have,

$\quad \sin x+\frac{1}{2} \cos x=\sin ^2\left(x+\frac{\pi}{4}\right)$

$\Rightarrow \quad 2 \sin x+\cos x=2\left[\sin \left(x+\frac{\pi}{4}\right)\right]^2$

$\Rightarrow 2 \sin x+\cos x$

$\qquad=2\left[\sin x \sin \frac{\pi}{4}+\cos x \cos \frac{\pi}{4}\right]^2$

$\Rightarrow \quad 2 \sin x+\cos x=2 \times \frac{1}{2}(\sin x+\cos x)^2$

$\Rightarrow \quad 2 \sin x+\cos x=\sin ^2 x+\cos ^2 x+2 \sin x$

$\Rightarrow \quad 2 \sin x+\cos x=1+2 \sin x \cos x$

$\Rightarrow \quad 2 \sin x-2 \sin x \cos x+\cos x-1=0$

$\Rightarrow \quad 2 \sin x(1-\cos x)-1(1-\cos x)=0$

$\Rightarrow \quad \quad(2 \sin x-1)(1-\cos x)=0$

$\Rightarrow \quad \sin x=\frac{1}{2} \text { and } \cos x=1$

$=x=\frac{\pi}{6}, \frac{5 \pi}{6} \text { and } x=0[\because x \in[0, \pi]]$

Sum of all $x$ is $\frac{\pi}{6}+\frac{5 \pi}{6}+0=\pi$